Number of $5$-card hands containing exactly two aces I have the above problem and I came up with the solution: 
$${4 \choose 2}{12 \choose 3}4^3$$
which is to choose $2$ aces from $4$ aces, choose $3$ values for the $3$ remaining and choose the suits for those $3$ cards.
However, another solution that I came up with is:
$${4 \choose 2}{48 \choose 3}$$
which is to choose $2$ aces from $4$ aces and choose the remaining $3$ cards out of remaining $48$.
These two solutions give different results, but I don't know which one is right. Can someone clarify it?
Thank you.
 A: I believe that what you calculated the first time, ${4 \choose 2}{12 \choose 3}4^3$, is the number of hands whose rank is "a pair of aces."
For example, 
$\{\spadesuit \mathrm A, \diamondsuit \mathrm A, \clubsuit 2, \spadesuit 8, \heartsuit \mathrm Q\}$
is a pair of aces.
You get such a hand if it contains two aces and three other cards, where none of the other three cards is the same value as any other card in the hand.
(I think what you call "value" is what many people call "rank".)
All three of your cards other than the aces will be different ranks because you chose three distinct ranks from the $12$ that are not aces.
At least, that is the only reasonable explanation I can easily see for choosing the factor $\binom{12}{3}$: choose $3$ ranks out of $12$ without repetition.
Then, since each of those three cards is a rank different from the other two, all four suits are available to choose from, hence you multiply by $4^3.$
The hand $\{\spadesuit \mathrm A, \diamondsuit \mathrm A, \clubsuit 8, \spadesuit 8, \heartsuit \mathrm Q\}$
would not be called "a pair of aces"; rather it would be called two pairs,
aces and eights.
It will not be counted among the ${4 \choose 2}{12 \choose 3}4^3$ hands in your first calculation because two $8$s and a Q are not three ranks chosen from among the $12$ ranks other than aces.
But I think you will agree that this hand also contains exactly two aces.
Your second way of counting accounts for "pair of aces" hands and also for other hands that contain exactly two aces, such as two pairs (a pair of aces and some other pair) or a full house (two aces and a three-of-a-kind).
A: It's always helpful to try to visualize the problem and how to approach it.
Think of the card arrangements as an array of length 5 named A.
Lets get the total number of ways that 2 slots of the 5 can be selected (Number of possible arrangements of 2 elements in an array of size 5) in A.
That's a simple combination. That is:
$${5 \choose 2}$$
Now lets get the total number of ways the available Aces can be arranged at those 2 selected slots in A.
$${4 \choose 2}$$
Multiplying both combinations given above gives us the number of ways 2 cards of a set of 4 cards can be placed at 5 slots:
$${5 \choose 2}{4 \choose 2}$$
NOTE: This is not the numbers of 5-card hands that has exactly 2 Aces. This is the total number of arrangements of 2 Aces of the 4 in A.
We have yet to compute the number of arrangements of the remaining cards.
We took out 4 cards (The Aces) of the 52 card deck so that leaves us with 48 remaining cards to be arranged at A and took 2 slots from A. So that gives us:
$${48 \choose 3}$$
So the answer is:
$${5 \choose 2}{4 \choose 2}{48 \choose 3}$$
A: For the correction of your first answer, consider all (three) cases of repetitions of value of cards
$(3,),$ e.g. Queen Queen Queen
$${12\choose1}{4\choose3}$$
$(2,1),$ e.g. Queen Queen King
$${12\choose2}{2\choose1}\cdot\left[{4\choose2}{4\choose1}\right]$$
$(1,1,1),$ e.g. Jack Queen King
$${12\choose3}\cdot\left[{4\choose1}+{3\choose2}{4\choose2}2!+{4\choose3}3!\right]$$
and 
$${48\choose3}=17296,$$
which, hopefully, is exactly the same as above, $48+3168+14080,$ and clearly ${48\choose3}$ is better/simpler.
